Question

If a, b, c and d are in G.P. show that (a ² + b ² + c ² ) (b 2 + c ² + d ² ) = (ab + bc + cd)².

Answer 1

It is given that a, b, c and d are in geometric progression (G.P.).

Let the common ratio of the G.P. be r.

Then we can write \[ b = ar,\quad c = ar^2,\quad d = ar^3. \]

Now consider the left-hand side (LHS):

\[ (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) \]

Substituting the values of b, c and d:

\[ (a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6) \]

Taking \( a^2 \) common from both brackets:

\[ a^2(1 + r^2 + r^4)\cdot a^2r^2(1 + r^2 + r^4) \]

\[ = a^4r^2(1 + r^2 + r^4)^2 \]

Now consider the right-hand side (RHS):

\[ (ab + bc + cd)^2 \]

Substituting the values:

\[ (a \cdot ar + ar \cdot ar^2 + ar^2 \cdot ar^3)^2 \]

\[ (a^2r + a^2r^3 + a^2r^5)^2 \]

Taking \( a^2r \) common:

\[ [a^2r(1 + r^2 + r^4)]^2 \]

\[ = a^4r^2(1 + r^2 + r^4)^2 \]

Thus, \[ (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2 \]

Hence, the given result is proved.